Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{4x^2 + 24x}{-3x^3 + 3x^2 + 270x} \div \dfrac{x + 1}{x^2 + 10x + 9} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{4x^2 + 24x}{-3x^3 + 3x^2 + 270x} \times \dfrac{x^2 + 10x + 9}{x + 1} $ First factor out any common factors. $p = \dfrac{4x(x + 6)}{-3x(x^2 - x - 90)} \times \dfrac{x^2 + 10x + 9}{x + 1} $ Then factor the quadratic expressions. $p = \dfrac {4x(x + 6)} {-3x(x + 9)(x - 10)} \times \dfrac {(x + 9)(x + 1)} {x + 1} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {4x(x + 6) \times (x + 9)(x + 1) } { -3x(x + 9)(x - 10) \times (x + 1)} $ $p = \dfrac {4x(x + 9)(x + 1)(x + 6)} {-3x(x + 9)(x - 10)(x + 1)} $ Notice that $(x + 9)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {4x\cancel{(x + 9)}(x + 1)(x + 6)} {-3x\cancel{(x + 9)}(x - 10)(x + 1)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $p = \dfrac {4x\cancel{(x + 9)}\cancel{(x + 1)}(x + 6)} {-3x\cancel{(x + 9)}(x - 10)\cancel{(x + 1)}} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $p = \dfrac {4x(x + 6)} {-3x(x - 10)} $ $ p = \dfrac{-4(x + 6)}{3(x - 10)}; x \neq -9; x \neq -1 $